Problem: The polar curve $r(\theta)=1-2\cos(\theta)$ is graphed for $0\leq\theta\leq\pi$. Let $R$ be the region in the third quadrant enclosed by the curve and the $x$ -axis. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{\scriptsize\dfrac{\pi}{3}}\left( \dfrac{1}{2}-2\cos(\theta)+2\cos^2(\theta)\right)d\theta$ (Choice B) B $ \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left( 1-4\cos(\theta)+4\cos^2(\theta)\right)d\theta$ (Choice C) C $ \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left( \dfrac{1}{2}-2\cos(\theta)+2\cos^2(\theta)\right)d\theta$ (Choice D) D $ \int_{0}^{\scriptsize\dfrac{\pi}{3}}\left( 1-4\cos(\theta)+4\cos^2(\theta)\right)d\theta$
Explanation: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. Notice that $r(0)=-1$, and indeed the curve starts at $(-1,0)$. So $\alpha=0$. $\beta$ is the $\theta$ -value in the range $0\leq\theta\leq\pi$ for which $r(\beta)=0$. So $\beta=\dfrac{\pi}{3}$. Let's plug ${r(\theta)=1-2\cos(\theta)}$, ${\alpha=0}$, and ${\beta=\dfrac{\pi}{3}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{0}}^{{\scriptsize\dfrac{\pi}{3}}}\dfrac{1}{2}\left({1-2\cos(\theta)}\right)^{2}d\theta \\\\ &= \int_{0}^{\scriptsize\dfrac{\pi}{3}}\dfrac{1}{2}\left( 1-4\cos(\theta)+4\cos^2(\theta)\right)d\theta \\\\ &= \int_{0}^{\scriptsize\dfrac{\pi}{3}}\left( \dfrac{1}{2}-2\cos(\theta)+2\cos^2(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{0}^{\scriptsize\dfrac{\pi}{3}}\left( \dfrac{1}{2}-2\cos(\theta)+2\cos^2(\theta)\right)d\theta$